Max K.E. of electron

Show that when a photon of energy E is scattered from a free electron at rest, the maximum kinetic energy of the recoiling electron is given by:
[tex]K_{max} = \frac{E^2}{E + mc^2/2}[/tex]

This a "Compton effect" problem.
The kinetic energy of the electron is given by:
[itex]K = E - hf'[/itex] where f' is the frequency of the scattered photon. But this doesn't even remotely resemble the result I need. Am I missing out on something?

Show that when a photon of energy E is scattered from a free electron at rest, the maximum kinetic energy of the recoiling electron is given by:
[tex]K_{max} = \frac{E^2}{E + mc^2/2}[/tex]

This a "Compton effect" problem.
The kinetic energy of the electron is given by:
[itex]K = E - hf'[/itex] where f' is the frequency of the scattered photon. But this doesn't even remotely resemble the result I need. Am I missing out on something?

You know that in the Compton effect, the wavelength of the scattered photon is longer than that of the incident photon as given by the equation : [tex]\Delta\lambda = \lambda_f - \lambda_i = \frac{h}{mc}(1-\cos\theta)[/tex] where the subscripts denote final and initial wavelength, theta is the angle of scatter and m is the rest electron mass.

By conservation of energy, the entire difference in photon energies between initial and final must be passed on to the electron as kinetic energy. So you want to maximise [tex]\Delta\lambda[/tex].

By looking at the constraints on cos theta, figure out the maximum possible value of [tex]\Delta\lambda[/tex] in terms of h, m and c.

Now express the wavelength in terms of photon energy E. Rearrange the equation until you've formed a relationship between the final and initial photon energies. The requried kinetic energy of the electron will be the difference between the initial and final photon energies.